Q56 / 56 · IAT (IISER) 2026 · Physics
Consider two Carnot engines of efficiencies $\eta_1$ and $\eta_2$. The first engine absorbs heat $Q_1$ from a heat reservoir $A$ and releases heat $Q_2$ to a heat reservoir $B$. The second engine takes heat $Q_2$ from $B$ and releases heat $Q_3$ to a heat reservoir $C$. If $Q_1 > Q_2 > Q_3$, what is the net efficiency of this combination of the two Carnot engines?
- Correct answerA$\eta_1 + \eta_2 - \eta_1\eta_2$
- B$\eta_1 + \eta_2 + \eta_1\eta_2$
- C$\eta_1\eta_2$
- D$\eta_1 + \eta_2$
Solution
The efficiency of the first Carnot engine is given by $\eta_1 = 1 - \dfrac{Q_2}{Q_1}$.
This can be rewritten as $\dfrac{Q_2}{Q_1} = 1 - \eta_1$.
The efficiency of the second Carnot engine is given by $\eta_2 = 1 - \dfrac{Q_3}{Q_2}$.
This can be rewritten as $\dfrac{Q_3}{Q_2} = 1 - \eta_2$.
The net efficiency of the combination of the two engines is the total work done divided by the total heat absorbed from the source.
Total work done is $W = W_1 + W_2 = (Q_1 - Q_2) + (Q_2 - Q_3) = Q_1 - Q_3$.
The net efficiency is $\eta = \dfrac{W}{Q_1} = \dfrac{Q_1 - Q_3}{Q_1} = 1 - \dfrac{Q_3}{Q_1}$.
Multiplying the ratios of heat gives $\dfrac{Q_3}{Q_1} = \dfrac{Q_3}{Q_2} \times \dfrac{Q_2}{Q_1} = (1 - \eta_2)(1 - \eta_1)$.
Substituting this into the expression for net efficiency:
$\eta = 1 - (1 - \eta_1)(1 - \eta_2)$
$\eta = 1 - (1 - \eta_1 - \eta_2 + \eta_1\eta_2)$
$\eta = \eta_1 + \eta_2 - \eta_1\eta_2$
Answer: $\eta_1 + \eta_2 - \eta_1\eta_2$
This can be rewritten as $\dfrac{Q_2}{Q_1} = 1 - \eta_1$.
The efficiency of the second Carnot engine is given by $\eta_2 = 1 - \dfrac{Q_3}{Q_2}$.
This can be rewritten as $\dfrac{Q_3}{Q_2} = 1 - \eta_2$.
The net efficiency of the combination of the two engines is the total work done divided by the total heat absorbed from the source.
Total work done is $W = W_1 + W_2 = (Q_1 - Q_2) + (Q_2 - Q_3) = Q_1 - Q_3$.
The net efficiency is $\eta = \dfrac{W}{Q_1} = \dfrac{Q_1 - Q_3}{Q_1} = 1 - \dfrac{Q_3}{Q_1}$.
Multiplying the ratios of heat gives $\dfrac{Q_3}{Q_1} = \dfrac{Q_3}{Q_2} \times \dfrac{Q_2}{Q_1} = (1 - \eta_2)(1 - \eta_1)$.
Substituting this into the expression for net efficiency:
$\eta = 1 - (1 - \eta_1)(1 - \eta_2)$
$\eta = 1 - (1 - \eta_1 - \eta_2 + \eta_1\eta_2)$
$\eta = \eta_1 + \eta_2 - \eta_1\eta_2$
Answer: $\eta_1 + \eta_2 - \eta_1\eta_2$
Hence, the correct answer is option: 'A'.